Well I think I have more or less an answer to my question. I have showed shown that the set of all maximal imprimitive transitive subgroups $H\leq S_N$ is of the form $$S_{N/r}^{r}\rtimes S_r$$ for $r|N$ and where $S_r$ acts by permutation on the coordinates of $S_{N/r}$. S_{N/r}^r$. So since I have an onto group homomorphism $$f:H\rightarrow S_n$$ I must conclude that$H\subseteq S_{2}^{n}\rtimes S_n$and that$H\supseteq S_n$. Finally, since I can produce an element$\tau\in H$that has a cycle of length larger than$n$which appears in its cycle presentation I may conclude that$H$is not contained in any maximal transitive imprimitive subgroups of$S_N$and therefore by maximality this implies that$H=S_N$. But this is absurd since it contradicts the imprimitivity of$H$. Therefore such an$H$does not exist. 1 Well I think I have more or less an answer to my question. I have showed that the set of all maximal imprimitive transitive subgroups$H\leq S_N$is of the form $$S_{N/r}^{r}\rtimes S_r$$ for$r|N$and where$S_r$acts by permutation on the coordinates of$S_{N/r}$. So since I have an onto group homomorphism $$f:H\rightarrow S_n$$ I must conclude that$H\supseteq S_n$. Finally, since I can produce an element$\tau\in H$that has a cycle of length larger than$n$which appears in its cycle presentation I may conclude that$H=S_N\$.